Consider the following electronic configuration of an element(P) : [Xe]4f14 5d1 6s2. Then correct st — Periodic Table and Periodicity Chemistry Question

💡 Solution & Explanation

Step 1: Identify the element 'P' with configuration [Xe] 4f14 5d1 6s2. This corresponds to Lutetium (Z=71), the last lanthanide. Step 2: The highest principal quantum number is n=6, indicating it is in the 6th period. Lanthanides and Lutetium are positioned in Group 3 of the d-block. Step 3: Conclude that element 'P' belongs to the 6th period and 3rd group, which is option (c).